\documentclass[12pt, a4paper, oneside]{ctexart}
\usepackage{amsmath, amsthm, amssymb, bm, color, framed, graphicx, hyperref, mathrsfs,esint}

\title{\vspace{-4cm}\textbf{河北师范大学数学分析真题}}
\author{宁鑫雨}
\date{\today}
\linespread{1.5}
\definecolor{shadecolor}{RGB}{241, 241, 255}
\newcounter{problemname}
\def\d{\mathrm{d}}
\newenvironment{problem}{\begin{shaded}\stepcounter{problemname}\noindent\textbf{题目\arabic{problemname}. }}{\end{shaded}}
\newenvironment{solution}{\par\noindent\textbf{解答. }}{\par}
\newenvironment{note}{\par\noindent\textbf{题目\arabic{problemname}的注记. }}{\par}
\pagestyle{plain}
\setlength{\parindent}{0pt}
\begin{document}
\date{}
\section*{2012年高等代数}
\begin{problem}[本题20分]
设$f_{1}(x),f_{2}(x)$是数域P上两个互素的多项式,$a_{1},a_{2}$是数域$P$中的两个数.证明:存在多项式$g(x)$满足
$$
g(x)=u_i(x)f_i(x)+a_i,i=1,2.
$$
\end{problem}
\begin{solution}
证明:\\
$\because$ $f_{1}(x)$,$f_{2}(x)$互素\\
$\therefore$ 存在$V_{1}(x)$,$V_{2}(x)$,使得$V_{1}(x)f_{1}(x)+V_{2}(x)f_{2}(x)=1$\\
$\therefore$ 令$V_{1}(x)=\frac{u_{1}(x)}{a_{2}-a_{1}}$ ,$V_{2}(x)=\frac{u_{2}(x)}{a_{1}-a_{2}}$\\
$\therefore$ $\frac{u_{1}(x)}{a_{2}-a_{1}}f_{1}(x)+\frac{u_{2}(x)}{a_{1}-a_{2}}f_{2}(x)=1$\\
$\therefore$ $u_{1}(x)f_{1}(x)-u_{2}(x)f_{2}(x)=a_{2}-a_{1}$\\
$\therefore$ $u_{1}(x)f_{1}(x)+a_{1}=u_{2}(x)f_{2}(x)+a_{2}$\\
令$g(x)=u_{1}(x)f_{1}(x)+a_{1}=u_{2}(x)f_{2}(x)+a_{2}$\\
$\therefore$ 原题得证
\end{solution}


\begin{problem}[本题20分]
证明：在实数域上,线性方程组$A^{T}A x =A^{T}\alpha$一定有解,这里$A$是$m \times n$的实矩阵,$\alpha$为任意$m$维的实向量.
\end{problem}
\begin{solution}
    $\because$ $A^{T}A x =0$,$A x =0$同解\\
    $\therefore$ $r(A^{T}A)=r(A)$\\
    又$\because$ $r(A^{T})=r(A)$
    $\therefore$ $r(A^{T}A)=r(A^{T})$\\
    $\therefore$ $r(A^{{T}}A:A^{{T}}\alpha)=r(A^{{T}}:A^{{T}}\alpha)=r(A^{{T}}:0)=r(A^{{T}})=r(A^{T}A)$\\
    即有$r(A^{{T}}A:A^{{T}}\alpha)=r(A^{T}A)$\\
    $\therefore$ $A^{T}A x =A^{T}\alpha$有解
\end{solution}


\begin{problem}[本题20分]
设$P^n$是数域$P$上的$n$维向量空间,证明：存在$P^n$的一个$r(0<r<n)$维子空间$W$,满足$W$中每个非零向量的个数均大于$n-r$.
\end{problem}
\begin{solution}
由题意可知$W=\{x\vert A x=0,r(A)=n-r\}$\\
构造$A=\left[
\begin{array}{c c c c}
    {{1}}&{{1}}&{{...}}&{{1}}\\ 
    {{y_{1}}}&{{y_{2}}}&{{...}}&{{y_{n}}}\\ 
    {{y_{1}}^{n-r-1}}&{{y_{2}}^{n-r-1}}&{{...}}&{{y_{n}}^{n-r-1}}\\ 
\end{array}\right]$,显然 $r(A)=n-r$\\
设$A=(\alpha_{1}\cdots\alpha_{n})$\\
则$Ax$=0,即$(\alpha_{1}\cdots\alpha_{n})\left(\begin{matrix}x_{1}\\ \vdots\\x_{n}\end{matrix} \right)=0$\\
反证假设$W$中每个非零向量的个数$\le n-r $\\
则$\alpha_{1}x_{1}+\cdots+\alpha_{n-r}\alpha_{n-r}=0$\\
又$\because$ $r(A)=n-r$\\
$\therefore$ $\alpha_{1}\cdots\alpha_{n-r}$线性无关\\
$\therefore$ $x_{1}=\cdots=x_{n-r}=0$\\
$\therefore$ 此向量均为零向量与非零向量矛盾\\
$\therefore$ 对于每个非零向量,非零向量的个数均大于$n-r$\\
\end{solution}

    
\begin{problem}[本题20分]
    设$\alpha,\beta$是n维欧氏空间中两个长度相等且线性无关的向量,证明：
    $$
    <\alpha+\beta>^{\bot}=\bigl(<\alpha>^{\bot}\bigcap<\beta>^{\bot}\bigr)+<\alpha-\beta>.
    $$
\end{problem}
\begin{solution}
    要证$ <\alpha+\beta>^{\bot}=\bigl(<\alpha>^{\bot}\bigcap<\beta>^{\bot}\bigr)+<\alpha-\beta>$\\
    % 只需证$ \bigl(<\alpha>^{\bot}\bigcap<\beta>^{\bot} \bigr)+<\alpha-\beta>-<\alpha-\beta>^\bot=\\
    % [\bigl(<\alpha>^{\bot}\bigcap<\beta>^{\bot} \bigr)<\alpha-\beta> ][\bigl(<\alpha>^{\bot}\bigcap<\beta>^{\bot} \bigr)<\alpha-\beta> ]\\
    % =0$\\
    只需证
    $\left( <\alpha>^\bot \bigcap <\beta>^\bot + <\alpha-\beta>\right) \cdot<\alpha+\beta>^\bot\\
    =\left[ \left( <\alpha>^\bot \bigcap <\beta>^\bot \right) \cdot <\alpha+\beta> \right] + \left[ <\alpha-\beta> \cdot <\alpha+\beta> \right]\\
    =0
    $\\
    设$\alpha=(a_{1},a_{2},\cdots,a_{n}),\beta=(b_{1},b_{2},\cdots b_{n})$\\
    则$\alpha-\beta=(a_{1}-b_{1},a_{2}-b_{2},\cdots a_{n}-b_{n})$,$\alpha+\beta=(a_{1}+b_{1},a_{2}+b_{2},\cdots a_{n}n+b_{n})$\\
    $\therefore$ 
    $<\alpha-\beta><\alpha+\beta>=a_{1}^{2}-b_{1}^{2}+a_{2}^{2}-b_{2}^{2}+\cdots+a_{n}^{2}-b_{n}^{2}=(a_1^{2}+a_{2}^{2}+\cdots+a_{n}^{2})-(b_{1}^{2}+b_{2}^{2}+\cdots+b_{n}^{2})$\\
    $\because$ $\alpha$,$\beta$长度相等,
    $\sqrt{<\alpha ,\alpha>}=\sqrt{a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}}=\sqrt{b_{1}^{2}+b_{2}^{2}+\cdots+b_{n}^{2}}=\sqrt{<\beta_{1},\beta_{2}>}$\\
    $\therefore$ $<\alpha-\beta><\alpha+\beta>=0$\\
    又$\because$ $\alpha,\beta$线性无关\\
    $\therefore$ $<\alpha>^{\bot}$与$<\beta>^{\bot}$也线性无关\\
    $\therefore$ $<\alpha>^{\bot}\bigcap<\beta>^{\bot}=\left\{ 0 \right\}$\\
    $\therefore$ $\bigl(<\alpha>^{\bot}\bigcap<\beta>^{\bot} \bigr)<\alpha+\beta>=0$\\
    $\therefore$ $[\bigl(<\alpha>^{\bot}\bigcap<\beta>^{\bot} \bigr)+<\alpha-\beta>]<\alpha+\beta>=0$\\
    $\therefore$ 原题得证
\end{solution}
    

\begin{problem}[本题30分]
    设$\eta$是欧氏空间$V$中一单位向量,定义
    $$
    \sigma(\alpha)=\alpha-2(\eta,\alpha)\eta.
    $$
    证明：\\
    1)$\sigma$是正交变换,这样的正交变换称为镜面反射;\\
    2)$\sigma$是第二类的;\\
    3)如果$n$维欧氏空间中,正交变换$\sigma$以1作为一个特征值,且属于特征值1的特征子空间$V_{1}$的维数为n-1,那么$\sigma$是镜面反射.
\end{problem}
\begin{solution}
    1)对欧氏空间中任意元素$\alpha$,$\beta$和实数$k_{1}$,$k_{2}$,有

    $
    \sigma(k_{1}\alpha+k_{2}\beta)=(k_{1}\alpha+k_{2}\beta)-2(\eta,k_{1}\alpha+k_{2}\beta)\eta\\
    =k_{1}\alpha+k_{2}\beta-2(\eta,k_{1}\alpha)\eta-2(\eta,k_{2}\beta)\eta\\
    =k_{1}[\alpha-2(\eta,\alpha)\eta]+k_{2}[\beta-2(\eta,\beta)\eta]\\
    =k_{1}\sigma(\alpha)+k_{2}\sigma(\beta)\\
    $
    $\therefore$$\sigma$是线性的\\
    又有
    $$
    (\sigma\alpha,\sigma\beta)=(\alpha-2(\eta,\alpha)\eta,\beta-2(\eta,\beta)\eta)\\
    =(\alpha,\beta)-2(\eta,\alpha)(\eta,\beta)-2(\eta,\beta)(\eta,\alpha)+4(\eta,\alpha)(\eta,\beta)(\eta,\eta)\\
    $$
    $\because$$(\eta,\eta)$=1\\
    $\therefore$ $(\sigma\alpha,\sigma\beta)=(\alpha,\beta)$\\
    $\therefore$ $\sigma$为正交变换\\
    2)由于$\eta$是单位向量，将它扩充为空间的一组标准正交基$\eta,\varepsilon_{2},\cdots\varepsilon_{n}$,则有\\
    $\sigma(\eta)=\eta-2(\eta,\eta)\eta=-\eta$\\
    $\sigma(\varepsilon_{i})=\varepsilon_{i}-2(\eta,\varepsilon_{i})\eta=\varepsilon_{i}$,i=2,$\cdots$,n \\
    $\therefore$ $\sigma(\eta,\varepsilon_{2},\cdots,\varepsilon_{n})=(\eta,\varepsilon_{2},\cdots,\varepsilon_{n})\left[\begin{matrix}-1& & & \\&1&&\\&&\ddots\\&&&1\end{matrix}\right]$\\
    $\therefore$ $\sigma$在基$\eta,\varepsilon_{2},\cdots,\varepsilon_{n}$下的矩阵$A$的行列式为-1\\
    $\therefore$ $\sigma$为第二类正交变换\\
    3)$\sigma$的特征值有$n$个,现已有$n-1$个1，另一个也为实数\\
    不妨设为$\lambda_{0}$,则存在一组基$\varepsilon_{1},\varepsilon_{2},\cdots,\varepsilon_{n}$,有\\
    $\sigma \varepsilon_{1} =\lambda_{0}\varepsilon_{1},\sigma \varepsilon_{i} =\varepsilon_{i},i=2,\cdots,n$\\
    $\because$$\sigma$是正交变换\\
    $\therefore$ $(\varepsilon_{1},\varepsilon_{1})=(\sigma\varepsilon_{1},\sigma\varepsilon_{1})=\lambda_{0}^{2}(\varepsilon_{1},\varepsilon_{1})$\\
    $\therefore$ $\lambda_{0}^{2}=1$\\
    $\because$ $V_{1}$是$n-1$维的\\
    $\therefore$ $\lambda_{0}=-1$\\
    $\therefore$ $\sigma \varepsilon_{1} =-\varepsilon_{1},\sigma \varepsilon_{i} =\varepsilon_{i},i=2,\cdots,n$\\
    $\because$$A$为实对称矩阵\\
    $\therefore$属于它的不同特征值的特征向量必正交\\
    $\therefore$ $(\varepsilon_{1},\varepsilon_{i})=0$,$(i=2,\cdots,n)$\\
    令 
    $\eta=\frac{\varepsilon_{1}}{|\varepsilon_{1}|}$,则$\eta$是与$\varepsilon_{2},\cdots,\varepsilon_{n}$正交的单位向量\\
    并且$\eta,\varepsilon_{2},\cdots,\varepsilon_{n}$组成一组基\\
    又有$\sigma\eta=\sigma(\frac{\varepsilon_{1}}{|\varepsilon_{1}|})=\frac{\sigma(\varepsilon_{1})}{|\varepsilon_{1}|}=-\frac{\varepsilon_{1}}{|\varepsilon_{1}|}=-\eta$\\
    任意$\alpha \in V $,$\alpha=k_{1}\eta+k_{2}\varepsilon_{2}+\ldots+k_{n}\varepsilon_{n}$\\
    有$(\alpha,\eta)=(k_{1}\eta+k_{2}\varepsilon_{2}+\ldots+k_{n}\varepsilon_{n},\eta)=k_{1}$\\
    $\sigma\alpha=k_{1}\sigma\eta+k_{2}\sigma\varepsilon_{2}+\ldots+k_{n}\sigma\varepsilon_{n}=\alpha-2(\alpha,\eta)\eta$\\
    $\therefore$ $\sigma$为镜面反射.
\end{solution}
    

\begin{problem}[本题40分]
    设矩阵
    $A=
    \begin{pmatrix}
        1&1&1\\ 
        1&2&1\\ 
        1&1&t
    \end{pmatrix}$,
    这里$t$是使得$A$为正定矩阵的最小正整数,试解下列各题:\\
    1)求$t$的值.\\
    2)求可逆矩阵$T$,使得$T^{-1}AT$为对角矩阵.\\
    3)求矩阵$L,D,U$,使得$A=LDU$,这里$L$为单位下三角矩阵,即对角线元素都是1的下三角矩阵,$U$为单位上三角矩阵,$D$为对角矩阵.\\
    4)设$W=\left\{X\middle|A X=X A,X\in R^{3\times3}\right\}$,求线性空间$W$的维数.
\end{problem}
\begin{solution}
    1)要使$A$为正定矩阵,所以$A$的所有顺序主子式会大于0\\
    $ |A|=\left|\begin{array}{l l l}{1}&{1}&{1}\\ {1}&{2}&{1}\\ {1}&{1}&{t}\end{array}\right|=2t-2>0$\\
    $\therefore$$t>1$\\
    $\because$ $t$为最小正整数\\
    $\therefore$$t=2$\\
    2)$|\lambda E-A|=
    \left|
        \begin{matrix}
            \lambda-1&-1&-1\\ 
            -1&\lambda-2&-1\\ 
            -1&-1&\lambda-2
        \end{matrix}
    \right|=
    (\lambda-1)(\lambda^{2}-4\lambda+1)=
    (\lambda-1)[\lambda-(2+\sqrt{3})][\lambda-(2-\sqrt{3})]$\\
    当$\lambda=1$时,可得到基础解系为$\left(\begin{array}{c}0\\ -1\\ 1\end{array}\right)$\\
    当$\lambda=2+\sqrt{3}$时,可得到基础解系为$\left(\begin{array}{c}\sqrt{3}-1\\ 1\\ 1\end{array}\right)$\\
    当$\lambda=2-\sqrt{3}$时,可得到基础解系为$\left(\begin{array}{c}-\sqrt{3}-1\\ 1\\ 1\end{array}\right)$\\
    $\therefore$可逆矩阵
    $T=\left[\begin{matrix}0&\sqrt{3}-1&-\sqrt{3}-1\\ {-1}&1&1\\ {-1}&1&1\end{matrix}\right]$,\\
    使得$T^{-1}AT=\left[\begin{matrix}1&&\\&2+\sqrt{3}&\\ &&2-\sqrt{3}\end{matrix}\right]$,\\
    对称矩阵对角化。\\
    3)设$L=\left[\begin{matrix}1&0&0\\ a&1&0\\ b&c&1\end{matrix}\right]$,
    $D=\left[\begin{matrix}\lambda_{1}&0&0\\ 0&\lambda_{2}&0\\ 0&0&\lambda_{3}\end{matrix}\right]$,
    $U=\left[\begin{array}{ccc}1&d&e\\ 0&1&f\\ 0&0&1\\ \end{array}\right]$\\
    则$\left[\begin{matrix}1&0&0\\ a&1&0\\ b&c&1\end{matrix}\right] \left[\begin{matrix}\lambda_{1}&0&0\\ 0&\lambda_{2}&0\\ 0&0&\lambda_{3}\end{matrix}\right]\left[\begin{array}{ccc}{{1}}&{{d}}&e\\ {{0}}&{{1}}&f\\ {{0}}&{{0}}&{{1}}\\ \end{array}\right]=
    \left[\begin{array}{c c c}{{1}}&{{1}}&{{1}}\\ {{1}}&{{2}}&{{1}}\\ {{1}}&{{1}}&{{2}}\\ \end{array}\right]$\\
    解得$a=b=d=e=1$,$c=f=0$,$\lambda_{1}=\lambda_{2}=\lambda_{3}=1$.\\
    $\therefore$
    $L=\left[\begin{matrix}1&0&0\\ 1&1&0\\ 1&0&1\end{matrix}\right]$,
    $D=\left[\begin{matrix}1&0&0\\ 0&1&0\\ 0&0&1\end{matrix}\right]$,
    $U=\left[\begin{array}{ccc}1&1&1\\ 0&1&0\\ 0&0&1\\ \end{array}\right]$即为所求\\
    4)设与$A$可交换的矩阵$X=\left[\begin{array}{c c c}{{a_{11}}}&{{a_{12}}}&{{a_{13}}}\\ {{a_{21}}}&{{a_{22}}}&{{a_{23}}}\\ {{a_{31}}}&{{a_{32}}}&{{a_{33}}}\\ \end{array}\right]$\\
    $AX=
    \left[\begin{array}{c c c}{{1}}&{{1}}&{{1}}\\ {{1}}&{{2}}&{{1}}\\ {{1}}&{{1}}&{{2}}\\ \end{array}\right]\left[\begin{array}{c c c}{{a_{11}}}&{{a_{12}}}&{{a_{13}}}\\ {{a_{21}}}&{{a_{22}}}&{{a_{23}}}\\ {{a_{31}}}&{{a_{32}}}&{{a_{33}}}\\ \end{array}\right]\\
    =\left[
        \begin{array}{c c c}
        {{a_{11}+a_{21}+a_{31}}}&{{a_{12}+a_{22}+a_{32}}}&{{a_{13}+a_{23}+a_{33}}}\\ 
        {{a_{11}+2a_{21}+a_{31}}}&{{a_{12}+2a_{22}+a_{32}}}&{{a_{13}+2a_{23}+a_{33}}}\\ 
        {{a_{11}+a_{21}+2a_{31}}}&{{a_{12}+a_{22}+2a_{32}}}&{{a_{13}+a_{23}+2a_{33}}}\\ 
        \end{array}
    \right]$\\
    $XA=
    \left[\begin{array}{c c c}{{a_{11}}}&{{a_{12}}}&{{a_{13}}}\\ {{a_{21}}}&{{a_{22}}}&{{a_{23}}}\\ {{a_{31}}}&{{a_{32}}}&{{a_{33}}}\\ \end{array}\right]\left[\begin{array}{c c c}{{1}}&{{1}}&{{1}}\\ {{1}}&{{2}}&{{1}}\\ {{1}}&{{1}}&{{2}}\\ \end{array}\right]\\
    =\left[
        \begin{array}{c c c}
        {{a_{11}+a_{12}+a_{13}}}&{{a_{11}+2a_{12}+a_{13}}}&{{a_{11}+a_{12}+2a_{13}}}\\ 
        {{a_{21}+a_{22}+a_{23}}}&{{a_{21}+2a_{22}+a_{23}}}&{{a_{21}+a_{22}+2a_{23}}}\\ 
        {{a_{31}+a_{32}+a_{33}}}&{{a_{31}+2a_{32}+a_{33}}}&{{a_{31}+a_{32}+2a_{33}}}\\ 
        \end{array}
    \right]$\\
    $\therefore$对应元素对应相等,则有
    $a_{23}=a_{32},a_{12}=a_{21}=a_{13}=a_{31},
    a_{22}=a_{33},a_{11}=a_{22}+a_{23}-2a_{21}$\\
    令$ a_{23}=a_{32}=c $,$a_{12}=a_{21}=a_{13}=a_{31}=a$,$a_{22}=y$\\
    $\therefore$
    $X=
    \left[
        \begin{array}{c c c}
            {{y+c-2a}}&{{a}}&{{a}}\\ 
            {{a}}&{{y}}&{{c}}\\ 
            {{a}}&{{c}}&{{y}}\\ 
        \end{array}
    \right]\\
    =y\left[\begin{array}{ccc}1&&\\&{{\frac{1}{3}}}\\ &&{{\frac{1}{3}}}\\ \end{array}\right]
    +2c\left[\begin{matrix}0&0&0\\ 0&0&\frac{1}{3}\\ 0&\frac{1}{3}&0\end{matrix}\right]
    +c\left[\begin{matrix}1&0&0\\ 0&0&\frac{1}{3}\\ 0&\frac{1}{3}&0\end{matrix}\right]
    +2a\left[\begin{matrix}0&{\frac{1}{3}}&{\frac{1}{3}}\\ {\frac{1}{3}}&0&0\\ {\frac{1}{3}}&0&0\\ \end{matrix}\right]
    +2y\left[\begin{array}{ccc}0&0&0\\ 0&\frac{1}{3}&0\\ 0&0&\frac{1}{3}\\ \end{array}\right]
    +a\left[\begin{matrix}{{-2}}&{{\frac{1}{3}}}&{{\frac{1}{3}}}\\ {{\frac{1}{3}}}&{{0}}&{{0}}\\ {{\frac{1}{3}}}&{{0}}&{{0}}\\ \end{matrix}\right]\\$
    $\therefore$$W$的维数为6
\end{solution}

\end{document}